Question

A stone is projected at an angle \(60^{\circ}\) to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at highest point of flight will be \(\left(\cos 60^{\circ}=0.5\right)\)

• A. \(1:4\)
• B. \(4:1\)
• C. \(2:1\)
• D. \(\sqrt{3}:2\)

Hint:

At the highest point only the horizontal velocity component remains, \(u\cos\theta\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At projection the speed is \(u\). At the highest point only the horizontal component \(u\cos 60^{\circ}\) remains.

Step 2: Kinetic Energies
\(K_1 = \tfrac12 m u^2\)
\(K_2 = \tfrac12 m\left(u\cos60^{\circ}\right)^2 = \tfrac12 m u^2\times\tfrac14\)

Step 3: Ratio
\(\dfrac{K_1}{K_2} = \dfrac{1}{1/4} = 4:1\)

Step 4: Compare the options
The ratio 4 : 1 is option B.

Final Answer:
The ratio of kinetic energies is 4 to 1. \[ \boxed{\text{(B) }4:1} \]